Equivalence of Definitions of Inverse Mapping

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Theorem

Let $S$ and $T$ be sets.


The following definitions of the concept of Inverse Mapping are equivalent:

Definition 1

Let $f: S \to T$ be a mapping.

Let $f^{-1} \subseteq T \times S$ be the inverse of $f$:

$f^{-1} := \set {\tuple {t, s}: \map f s = t}$


Let $f^{-1}$ itself be a mapping:

$\forall y \in T: \tuple {y, x_1} \in f^{-1} \land \tuple {y, x_2} \in f^{-1} \implies x_1 = x_2$

and

$\forall y \in T: \exists x \in S: \tuple {y, x} \in f$


Then $f^{-1}$ is called the inverse mapping of $f$.

Definition 2

Let $f: S \to T$ and $g: T \to S$ be mappings.

Let:

$g \circ f = I_S$
$f \circ g = I_T$

where:

$g \circ f$ and $f \circ g$ denotes the composition of $f$ with $g$ in either order
$I_S$ and $I_T$ denote the identity mappings on $S$ and $T$ respectively.

That is, $f$ and $g$ are both left inverse mappings and right inverse mappings of each other.


Then:

$g$ is the inverse (mapping) of $f$
$f$ is the inverse (mapping) of $g$.


Proof

Definition 1 implies Definition 2

Let $f^{-1}: T \to S$ be an inverse mapping of $f: S \to T$ by definition 1.

From Mapping is Injection and Surjection iff Inverse is Mapping it follows that $f^{-1}$ is a bijection.

By Composite of Bijection with Inverse is Identity Mapping:

$f^{-1} \circ f = I_S$
$f \circ f^{-1} = I_T$


Thus $f^{-1}: T \to S$ is an inverse mapping of $f: S \to T$ by definition 2.

$\Box$


Definition 2 implies Definition 1

Let $g: T \to S$ be an inverse mapping of $f: S \to T$ by definition 2.

From Left and Right Inverse Mappings Implies Bijection it follows that both $f$ and $g$ are bijections.

From Mapping is Injection and Surjection iff Inverse is Mapping it follows that $g = f^{-1}$.

Thus $g: T \to S$ is an inverse mapping of $f: S \to T$ by definition 1.

$\blacksquare$


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